Calculate the pH of 10-8 M HCl .( log 11=1.0414)

Question

Calculate the pH of 10-8 M HCl .( log 11=1.0414) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/cdd179018bd5a5f66417a3e359ac1a66-.png" /> [ H +] text total =(10-8+10-7) M =10-8(1+10) =11 × 10-8 (Neglecting common -ion effect ) pH = log ( H +)=- log (11 × 10-8) =6.96=7

Q. Calculate the $p H$ of $1 0^{- 8} M H Cl . (lo g 11 = 1.0414)$

$[H^{+}]_{total} = (1 0^{- 8} + 1 0^{- 7}) M = 1 0^{- 8} (1 + 10)$
$= 11 \times 1 0^{- 8}$ (Neglecting common -ion effect )
$p H = lo g (H^{+}) = - lo g (11 \times 1 0^{- 8})$
$= 6.96 = 7$

Q. Calculate the pH of 10−8MHCl.(log11=1.0414)

[H+]total ​=(10−8+10−7)M=10−8(1+10) =11×10−8 (Neglecting common -ion effect ) pH=log(H+)=−log(11×10−8) =6.96=7

Q. Calculate the $p H$ of $1 0^{- 8} M H Cl . (lo g 11 = 1.0414)$

$[H^{+}]_{total} = (1 0^{- 8} + 1 0^{- 7}) M = 1 0^{- 8} (1 + 10)$
$= 11 \times 1 0^{- 8}$ (Neglecting common -ion effect )
$p H = lo g (H^{+}) = - lo g (11 \times 1 0^{- 8})$
$= 6.96 = 7$